consider a hollow hemisphere electric field at centre is sigma/4e consider heisphere of 3 parts each of 60 degrees now, 2E sin30 + E = sigma/4e E=sigma/8e sigma=1
Assuming 3 slices so that angle at center is 180. Calculating field by integration its one component will cancel out and it is 1/2e (made a silly mistake previous time). For one slice due to symmetry it is 1/6e.
Bhaiya please explain that conic one.i did not get it.
dunno smtyms u jst find ways 2 evade pen n paper..(pen ppr se to hemisphere ki field nikalni padti na..!)..yeah i missed the vector nature..will take care next time onwards..
1/2(pi)e ?
ReplyDeletee=epsilon not..(agar ki hogi to calculation mistake hi ki hogi..)
This comment has been removed by the author.
ReplyDeletesigma/2(pi)(epsilon)
ReplyDeletesigma=surface charge density
1/4(pi)e?
ReplyDeletee=epsilon not
none of you got it right...
ReplyDeleteand i can't understand how you all got 'pi's in your answers...
first find the field at the centre of a semicircle
ReplyDeleteit comes 2kQ/(pi)R^2 k=1/4(pi)(epsilon)
then consider a small half-ring at an angle A with the horizontal
let P be the surface charge density
then 2kP(pi)(R)(R dA)/(pi)R^2 = 2kP dA
now we take the sin and cos components and add them vectorially.
E(x)=integral 0 to pi/3 2kP cosA dA
E(y)=integral 0 to pi/3 2kP sinA dA
squaring and then taking root
E = 2kP = P/2(pi)(epsilon)
what's wrong ??
a hemisphere can't be assumed to be made up of rings as you have described.
ReplyDelete1/6e ?
ReplyDeletei think you are on the right track..yet, your answer is quite wrong...
ReplyDelete1/4e
ReplyDelete1/6e sounded better.
ReplyDeletewhats your feild at the center of a uniformly charged hollow sphere of surface charge density 1???
is it 1/8e
ReplyDeletehollow sphere k center pe to zero hi hoga..i used newtons third law to calculate it..
ReplyDeletelets say a charge q is placed a the center
F(on sphere due to q)=k(sigma)[(4pi r2)/6 ]*q/r2
now F(on sphere due to q)=F(on q due to sphere)=q*E(due to sphere at center)
so E=sigma/6e
(here sphere refers to that watermelon slice)
consider a hollow hemisphere
ReplyDeleteelectric field at centre is sigma/4e
consider heisphere of 3 parts each of 60 degrees
now, 2E sin30 + E = sigma/4e
E=sigma/8e
sigma=1
E=1/8e
Assuming 3 slices so that angle at center is 180.
ReplyDeleteCalculating field by integration its one component will cancel out and it is 1/2e (made a silly mistake previous time).
For one slice due to symmetry it is 1/6e.
Bhaiya please explain that conic one.i did not get it.
i meant hollow 'hemisphere' in my last comment..
ReplyDeleteKapil's latest soln. is perfect. Arun, you missed out the vector nature of fields, Digvijay, you too...
The force on the spherical region wont be as easy to calculate...
dunno smtyms u jst find ways 2 evade pen n paper..(pen ppr se to hemisphere ki field nikalni padti na..!)..yeah i missed the vector nature..will take care next time onwards..
ReplyDelete